Question

What is the equation of the circle that has its center at -26,120 and passed through the origin

Answer 1

well, first off let's check those two points, we know it's centerd at (-26 , 120) and we also know it passes through (0 , 0), so the distance between those two points is its radius

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{radius}{r}=√((~~-26 - 0~~)^2 + (~~120 - 0~~)^2) \implies r=√((-26)^2 + (120 )^2) \\\\\\ r=√(( -26 )^2 + ( 120 )^2) \implies r=√( 676 + 14400 ) \implies r=√( 15076 ) \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{√(15076)}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(√(15076))^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill`